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Prove that the derivative of tanh x is sech2x

Prove the following formula for the derivative of the hyperbolic tangent,

    \[ D(\tanh x) = \operatorname{sech}^2 x. \]


Proof. We know from the previous two exercises (here and here that

    \[ D(\sinh x) = \cosh x \qquad \text{and} \qquad D(\cosh x) = \sinh x. \]

Furthermore, we know from this exercise that

    \[ \cosh^2 x - \sinh^2 x = 1. \]

Thus, we can compute

    \begin{align*}  D(\tanh x) &= D \left( \frac{\sinh x}{\cosh x} \right) \\  &= \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} \\  &= \frac{1}{\cosh^2 x} \\  &= \operatorname{sech}^2 x. \qquad \blacksquare \end{align*}

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